Question

The digit at the tens place of a two-digit number is three times the digit at ones place. If the sum of this
number and the number formed by reversing its digits is 88, find the number.​

Answer 1

Answer:

Let the digit at tens place be x and the digit in ones place be y.

Then the original number will be 10x+y

The digit at tens place is three times the digit at the units place.

i.e. x=3y ..(a)

The sum of this number and the number formed by reversing its digits is 88.

(10x+y)+(10y+x)=88

$⇒11x+11y=88 \\ \\ ⇒x + y = 8..(b)$

Substitute the value of equation (a) in equation (b).

Therefore, 3y+y=8

$⇒4y = 8 \\ \\ ⇒y = 2$

Now, x=3y=3×2=6

Therefore, The number is 62.